Question: In a class of $8$, there are $3$ students who forgot their lunch. If the teacher chooses $4$ students, what is the probability that none of the four of them forgot their lunch?
We can think about this problem as the probability of $4$ events happening. The first event is the teacher choosing one student who remembered his lunch. The second event is the teacher choosing another student who remembered his lunch, given that the teacher already chose someone who remembered his lunch, and so on. The probabilty that the teacher will choose someone who remembered his lunch is the number of students who remembered their lunch divided by the total number of students: $\dfrac{5} {8}$ Once the teacher's chosen one student, there are only $7$ left. There's also one fewer student who remembered his lunch, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also remembered his lunch is $\dfrac{4} {7}$ The probability of the teacher picking two students who remembered their lunch must then be $\dfrac{5} {8} \cdot \dfrac{4} {7}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $4$ students such that none of them forgot their lunch is $\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}\cdot\dfrac{2}{5} = \dfrac{1}{14}$